If the roots of the equation 3x^3-26x^2+52x-2 | vedclass

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(A) Let the roots of the equation $3x^3-26x^2+52x-24=0$ be $\frac{a}{r}, a, ar$.We know that for a cubic equation $Ax^3+Bx^2+Cx+D=0$,the product of th

Vedclass · Accepted Answer

$\frac{8}{3}$

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(A) Let the roots of the equation $3x^3-26x^2+52x-24=0$ be $\frac{a}{r}, a, ar$.We know that for a cubic equation $Ax^3+Bx^2+Cx+D=0$,the product of the roots is $-\frac{D}{A}$.Thus,$\frac{a}{r} 	imes a 	imes ar = -\frac{(-24)}{3} = 8$.$a^3 = 8 \Rightarrow a = 2$.Since $a=2$ is a root,it must satisfy the equation: $3(2)^3 - 26(2)^2 + 52(2) - 24 = 24 - 104 + 104 - 24 = 0$.Now,the sum of the roots is $\frac{a}{r} + a + ar = -\frac{(-26)}{3} = \frac{26}{3}$.Substituting $a=2$: $\frac{2}{r} + 2 + 2r = \frac{26}{3} \Rightarrow \frac{2}{r} + 2r = \frac{26}{3} - 2 = \frac{20}{3}$.Dividing by $2$: $\frac{1}{r} + r = \frac{10}{3} \Rightarrow 3r^2 - 10r + 3 = 0$.Solving for $r$: $(3r-1)(r-3) = 0$,so $r = 3$ or $r = \frac{1}{3}$.The roots are $\frac{2}{3}, 2, 6$.The possible sums of two roots are $\frac{2}{3}+2 = \frac{8}{3}$,$2+6 = 8$,and $\frac{2}{3}+6 = \frac{20}{3}$.Comparing with the options,the correct sum is $\frac{8}{3}$.

If the roots of the equation $3x^3-26x^2+52x-24=0$ are in geometric progression,then the sum of two of its roots is

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